纵向数据或称集团数据、面板数据, 经常出现在生物、医学、经济和社会科学的研究中. 它是对同同一个个体进行多次观测, 所得数据是相关的, 但相关系数未知. 广义估计方程(GEE)^[1]是分析响应变量是离散或非负的纵向数据回归问题的重要方法. GEE方法的一个显著特点是即使每个个体多次观测值之间的相关系数(方差)假定错误, 只要均值函数假定正确, 所得回归系数的GEE估计仍具有相合性和渐近正态性. 若相关系数假定也正确, 则得到的GEE估计方差最小.

设$ Y_{ij} $是第$ i $个个体的第$ j $次观测响应变量的值, 回归因子$ X_{ij} $是$ p\times 1 $设计阵, $ i = 1, 2, \cdots , n $, $ j = 1, 2, \cdots , m_i $. 记$ Y_i = (Y_{i1}, \cdots , Y_{im_i})^\tau $, $ X_i = (X_{i1}, \cdots , X_{im_i}) $, 单调递增的$ \sigma $代数$ {{\cal F}}_{i}: = \sigma\{(Y_k, X_k), k = 1, \cdots, i\} $. 设$ X_{i} $是$ {\cal F}_{i-1} $可测, 即$ X_{i} $的设计依赖于前面的观察值, 称为自适应的, 且假设

(1.1) $ \begin{equation} \mu_{ij}(\beta): = \mbox{E}(Y_{ij}|{{\cal F}}_{i-1}) = \mu(X_{ij}^\tau\beta), \end{equation} $

(1.2) $ \begin{equation} \sigma_{ij}(\beta): = \mbox{Var}(Y_{ij}|{{\cal F}}_{i-1}) = \dot\mu(X_{ij}^\tau\beta), \end{equation} $

其中'$ \tau $'表示矩阵的转置, $ \mu(\theta) $称为联系函数(link), $ \dot\mu(\theta)>0 $是它的导数, $ \beta $是回归参数向量. 若$ \mu(\theta ) = \theta $, 就得到线性回归模型, $ \mu(\theta ) = e^\theta/(1+e^\theta) $就得到分析二分类数据的logit模型, $ \mu(\theta ) = e^\theta $就得到分析计数数据的对数线性Poisson模型. 记$ \mu_i(\beta) = (\mu_{i1}(\beta), \cdots , \mu_{im_i}(\beta))^\tau $, $ A_i(\beta) = \mbox{diag}(\sigma_{i1}(\beta), \cdots , \sigma_{im_i}(\beta)) $, 其中diag($ v $) 表示对角矩阵, 其对角线元素是向量$ v $的元素. 文献[2]和文献[3]定义GEE估计$ \hat\beta_n $为下列方程的根

(1.3) $ \begin{eqnarray} g_n(\beta) = \sum\limits_{i = 1}^nX_iA_i^{1/2}(\beta) R_i^{-1} A_i^{-1/2}(\beta)(Y_i-\mu_i(\beta)) = 0, \end{eqnarray} $

其中$ R_i $是工作相关阵.

GEE是广义线性模型(GLM)的推广, GLM和GEE的建模可参看文献[1, 4-7]. GLM大样本性质的可参看文献[8-13]. GEE大样本性质的可参看文献[2, 3, 14]. 文献[2]严格证明了工作相关阵已知, 每个个体观测次数可以无界情形下GEE估计的大样本性质. 文献[14]证明了协变量的维数可以无界情形下GEE估计的渐近性质. 文献[3]证明了具有估计工作相关阵, 误差是鞅差情形GEE估计的相合性和渐近正态性. 文献[2, 14]都假设$ Y_i, i\ge 1 $是相互独立的, $ X_{ij} $是固定的. 文献[3]假设$ \mbox{E}(Y_{ij}|{{\cal F}}_{i-1}) = \mu(X_{ij}^\tau\beta) $, 而设计阵$ X_{ij} $也还是假设是固定的, 非随机的. 自适应设计在心里学研究, 控制理论有广泛地应用, 可参见文献[6, 7, 15, 16]. 该文在较弱的条件下证明了自适应设计GEE回归参数估计的渐近性质, 并做了数值模拟. 把文献[2]和[3]的相应结果推广到了设计阵是自适应情形, 并且对Fisher信息阵的特征根的要求降到了最低.

在本文中, 函数在参数$ \beta $真值$ \beta_0 $点取值, 经常省略$ \beta_0 $, 如记$ A_i = A_i(\beta_0), \mu_i = \mu_i(\beta_0) $. $ C, C_1, C_2, \ldots $代表与$ n $无关的正常数, 在不同地方可以表示不同值. 为了得到我们的主要结果, 需要如下假设条件:

(A1) (ⅰ) $ \|X\|_n^2 / \lambda_{\min}(F_n)\to_{a.s} 0, $ (ⅱ) $ \sup\limits_{i\ge 1}m_i<\infty $, 其中Fisher信息阵$ F_n = $$ \sum\limits_{i = 1}^n X_iX_i^\tau $, $ \|X\|_n = \max\limits_{ i\le n, j\le m_i} \|X_{ij}\|^2 $, 对一个矩阵$ T $, $ \|T\| = [\mbox{trace}(TT^\tau)]^{1/2} $, $ \lambda_{\min}(T) $和$ \lambda_{\max}(T) $分别表示矩阵$ T $的最小和最大特征值.

(A2) 未知参数$ \beta $属于紧集$ {\mathbb B}\subset {{\Bbb R}} ^{p} $, 参数真值$ \beta_{0} $是$ \mathbb B $的内点.

(A3) 存在一个常数$ \alpha_0>2 $使得

$ \sup\limits_{i\ge 1, j\le m_i}E (\|Y^*_{ij} \|^{\alpha_0}|{\cal F}_{i-1})<\infty{\quad} a.s., $

其中$ Y_{ij}^* = Y_{ij}^*(\beta_0), Y_{ij}^*(\beta) = A_{ij}^{-1/2}(\beta)(Y_{ij}-\mu_{ij}(\beta)) $.

(A4) 概率为1, $ \inf\limits_{i\ge1}\lambda_{\min}( R_i)>0 $, $ \inf\limits_{i\ge1}\lambda_{\min}(\bar R_i)>0 $, 其中$ \bar R_i = E(Y_i^*(Y_i^*)^\tau|{\cal F}_{i-1}) $是真实相关阵, $ Y_{i}^* = Y_{i}^*(\beta_0), Y_{i}^*(\beta) = A_{i}^{-1/2}(\beta)(Y_{i}-\mu_{i}(\beta)). $

(A5) 存在一个非随机矩阵$ \bar M_n $使得$ \bar M_n^{-1}M_n^{1/2}\to_p I_p $, 其中$ I_p $是$ p $阶单位阵, $ M_n = M_n(\beta_0), $$ M_n(\beta) = \sum\limits_{i = 1}^n X_iA_i^{1/2} R_i^{-1}\bar R_i R_i^{-1}A_i^{1/2}X_i^\tau $.

(A6) (ⅰ) $ \sup\limits_{n\ge 1}\sup\limits_{i\le n, j\le m_i, \beta\in \tilde B_n(r)} |\mu^{(k)}(X_{ij}^\tau\beta)|<\infty \; \; a.s. $, $ \mu^{(k)} $表示$ \mu $的$ k $阶导数, $ k = 1, 2, 3 $. (ⅱ) $ \inf\limits_{n\ge 1}\inf\limits_{i\le n, j\le m_i, \beta\in \tilde B_n(r)}\dot \mu(X_{ij}^\tau\beta)>0\; \; a.s., $其中$ \tilde B_n(r) = \{\beta :\|\bar M_n^\tau(\beta-\beta_0)\|\le r\} $, $ r $是任意正常数.

定理2.1  若假设(A1)–(A6)成立. 则存在一个随机变量序列$ \{\hat\beta_n\} $使得

(2.1) $ \begin{equation} P( g_n(\hat\beta_n) = 0)\to 1, \qquad \hat\beta_n\to_p\beta_{0}, \nonumber \end{equation} $

而且

(2.2) $ \begin{equation} M_n^{-1/2} H_n(\hat\beta_n-\beta_{0})\to_d N(0, I_p), \nonumber \end{equation} $

其中

$ H_n = H_n(\beta_0), H_n(\beta) = \sum\limits_{i = 1}^nX_iA_{i}^{1/2}(\beta) R_i^{-1}A_{i}^{1/2}(\beta)X^\tau_i. $

注2.1  对纵向数据, 文献[2]研究的设计阵$ X_{ij} $是固定的, 文献[3]的$ X_{ij} $也是非随机的, 只是误差是鞅差. 显然都是我们自适应设计的特殊情形. 而且文献[3]要求$ n^{1/2}\|X\|_n^2/\underline\lambda_n\to 0, $其中$ \underline\lambda_n: = \lambda_{\min}(F_n) $. 对非纵向数据($ m_i = 1, i\ge 1) $, 自适应情形, 当$ \sup\limits_{n\ge 1}\|X\|_n<\infty \; \; a.s. $时, 文献[17]也要求$ \underline\lambda_n\ge C n^{1/2}\; a.s. $. 而我们只需要$ \underline\lambda_n\to\infty\; a.s. $, 这应该是相合性和渐近正态性需要的最弱条件, 因为对线性模型的最小二乘估计, 此条件是相合的必要条件.

注2.2  条件(A1)(ⅰ)类似Lindeberg-Feller中心极限定理中的Feller条件. 对自适应设计阵情形, 假设(A5)是需要的, 可参看文献[11, 15, 17]. 文献[15]举了一个反例, 说明若没有稳定条件(A5), 则线性模型最小二乘估计的渐近正态性不成立.

注2.3  假设(A4)和(A6)(ⅱ) 要求随机变量(协)方差有界, 而且不趋于0. (A3)是中心极限定理的李雅普诺夫条件. (A6)(ⅰ)是联系函数(link) 的光滑性条件. 这些都是文献中常用条件. 对纵向数据固定设计阵特殊情形, 假设(A2), (A3), (A4)和(A6), 可参看文献[2, 3, 14]. 对非纵向数据这种特殊情形, 文献[9-13, 16, 17]都假设$ \sup\limits_{ij}|X_{ij}|<\infty $ a.s.等条件成立, 很容易推出满足我们的条件(A6). 线性模型和logit模型, 假设(A6)(ⅰ)显然成立; 对数线性Poisson模型, 假设(A6)(ⅰ) 简化为$ \dot \mu(X_{ij}^\tau\beta) $一致有界a.s., 因为$ \dot\mu = \mu^{(2)} = \mu^{(3)} $.

为了证明定理2.1, 我们需要如下一些引理.

引理3.1^[18]  设$ C $是$ {{\Bbb R}} ^n $中的有界开集, 记$ \bar C $和$ \partial C $分别为$ C $的闭包和边界. 若对某个$ x_0\in C $和所有$ x\in \partial C $, 连续映射$ F: \bar C\to {{\Bbb R}} ^n $满足$ (x-x_0)^\tau F(x)\le 0 $. 则$ F(x) = 0 $在$ \bar C $中有一个根.

引理3.2^[19]  设$ \{S_{ni}, {\cal F}_{ni}, 1\le i\le k_n, n\ge 1\} $是零均值, 平方可积鞅, 其鞅差记为$ X_{ni} $, $ \eta^2 $是一个几乎处处有限的随机变量. 若对任意$ \epsilon>0 $, 有

$ \sum\limits_{i = 1}^{k_n }E(X_{ni}^2I(|X_{ni}|\ge\epsilon)|{\cal F}_{n, i-1})\to_p 0, \qquad \sum\limits_{i = 1}^{k_n }E(X_{ni}^2 |{\cal F}_{n, i-1})\to_p \eta^2, \nonumber $

则

$ S_{nk_n} = \sum\limits_{i = 1}^{k_n } X_{ni}\to_d Z, \nonumber $

其中随机变量$ Z $有特征函数$ E \exp(-\eta^2t^2/2) $, $ I(\cdot) $表示集合的示性函数, $ \sigma $ -代数: $ {\cal F}_{n, i-1}\subseteq{\cal F}_{n, i}, \;1\le i\le k_n, n\ge 1 $.

引理3.3^[2]  记$ D_n(\beta): = -\frac{\partial}{\partial\beta^\tau} g_n(\beta) $, 则

$ D_n(\beta) = H_n(\beta)- B^{[1]}_n(\beta)- B^{[2]}_n(\beta)- {\cal E}^{[1]}_n(\beta)-{\cal E}^{[2]}_n(\beta), \nonumber $

其中

$ \begin{eqnarray} &&B^{[1]}_n(\beta) = \frac{-1}{2}\sum\limits_{i = 1}^nX_i \mbox{diag}[ R_i^{-1}A_i^{-1/2}(\beta)(\mu_i-\mu_i(\beta))]A_i^{-1/2}(\beta)\ddot\mu_i(\beta)X_i^\tau, \\ && B^{[2]}_n(\beta) = \frac{1}{2}\sum\limits_{i = 1}^nX_iA_i^{1/2}(\beta) R_i^{-1}\mbox{diag}[\mu_i-\mu_i(\beta)]A_i^{-3/2}(\beta)\ddot\mu_i(\beta)X_i^\tau, \\ && {\cal E}^{[1]}_n(\beta) = \frac{-1}{2}\sum\limits_{i = 1}^nX_i \mbox{diag}[ R_i^{-1}A_i^{-1/2}(\beta)(Y_i-\mu_i)]A_i^{-1/2}(\beta)\ddot\mu_i(\beta)X_i^\tau, \\ && {\cal E}^{[2]}_n(\beta) = \frac{1}{2}\sum\limits_{i = 1}^nX_iA_i^{1/2}(\beta) R_i^{-1}\mbox{diag}[Y_i-\mu_i]A_i^{-3/2}(\beta)\ddot\mu_i(\beta)X_i^\tau, \\ &&\ddot\mu_i(\beta) = \mbox{diag}(\ddot\mu_{i1}(\beta), \cdots, \ddot\mu_{im}(\beta)) = \mbox{diag}(\ddot\mu(X^\tau_{i1}\beta), \cdots, \ddot\mu(X^\tau_{i1}\beta)), \end{eqnarray} $

$ \ddot \mu $是$ \mu $的二阶导数.

下面引理3.4至引理3.7的证明放在后面第5节.

引理3.4  若假设(A1)–(A6)成立, 则

$ \begin{eqnarray} \sup\limits_{\beta\in \tilde B_n(r)}\|H_n^{-1/2} H_n(\beta)H_n^{-1/2}-I_p\|\to_p0. \end{eqnarray} $

引理3.5  若假设(A1)–(A6)成立, 则

$ \begin{eqnarray} \sup\limits_{\beta\in \tilde B_n(r)}\|H_n^{-1/2} B^{[1]}_n(\beta)H_n^{-1/2}\|\to_p0, \\ \sup\limits_{\beta\in \tilde B_n(r)}\|H_n^{-1/2} B^{[2]}_n(\beta)H_n^{-1/2}\|\to_p0. \end{eqnarray} $

引理3.6  若假设(A1)–(A6)成立, 则

$ \begin{eqnarray} \sup\limits_{\beta\in \tilde B_n(r)}\|H_n^{-1/2} {\cal E}^{[1]}_n(\beta)H_n^{-1/2}\|\to_p0, \\ \sup\limits_{\beta\in \tilde B_n(r)}\|H_n^{-1/2} {\cal E}^{[2]}_n(\beta)H_n^{-1/2}\|\to_p0. \end{eqnarray} $

引理3.7  若假设(A1)–(A6)成立, 则

$ \begin{eqnarray} M_n^{-1/2} g_n(\beta_0) \to_d N(0, I). \end{eqnarray} $

定理2.1的证明  由引理3.3至3.6可得

(3.1) $ \begin{eqnarray} \sup\limits_{\beta\in \tilde B_n(r)}\|H_{n}^{-1/2}D_{n}(\beta)H_{n}^{-1/2}-I\| = o_p(1). \end{eqnarray} $

由式(3.1) 和$ \|\int \cdot\|\le\int\|\cdot\| $知

(3.2) $ \begin{eqnarray} \sup\limits_{\beta\in \tilde B_n(r)}\|H_{n}^{-1/2}D^*_{n}(\beta)H_{n}^{-1/2}-I\| = o_p(1), \end{eqnarray} $

其中$ D^*_{n}(\beta) = \int_0^1 D_{n}(\beta_0+t\beta){\rm d}t $, 积分是对矩阵每个元素进行积分. 由式(3.2)知

(3.3) $ \begin{eqnarray} \sup\limits_{\beta\in \tilde B_n(r)}\left\|\left[H_{n}^{-1/2}D^*_{n}(\beta)H_{n}^{-1/2}\right]^{-1}-I\right\| = o_p(1). \end{eqnarray} $

由假设条件(A4)至(A6), 可得对任意$ \epsilon>0 $, 存在$ C, C_1, C_2, C_3 $使

(3.4) $ \begin{eqnarray} P(CI\le R_i\le C_1\bar R_i\le C_2A_i\le C_3I, i\ge 1)>1-\epsilon, \end{eqnarray} $

其中$ I $是单位阵, 且

(3.5) $ \begin{eqnarray} P(F_n\le CH_n\le C_1M_n\le C_2\bar M_n\bar M_n^\tau\le C_3F_n)>1-\epsilon. \end{eqnarray} $

由引理3.7和式(3.5)知

(3.6) $ \begin{eqnarray} \|H_{n}^{-1/2}g_{n}\| = \|H_{n}^{-1/2}M_{n}^{1/2}\times M_{n}^{-1/2}g_{n}\| = O_p(1). \end{eqnarray} $

记$ \partial \tilde B_n(r) = \{\beta :\|\bar M_n^\tau(\beta-\beta_0)\| = r\} $. 由式(3.5)知, 对任意$ \epsilon>0 $, 存在$ C_5>0 $使得

(3.7) $ \begin{eqnarray} P\Big(\inf\limits_{\beta\in \partial \tilde B_n(r)}(\beta-\beta_0)^\tau H_n(\beta-\beta_0)\ge C_5r^2\Big)>1-\epsilon, \end{eqnarray} $

由式(3.2), 对任意$ \epsilon>0 $, 当$ n $充分大时

(3.8) $ \begin{eqnarray} P\Big(\inf\limits_{\|\lambda\| = 1, \beta\in \tilde B_n(r)}\lambda^\tau H^{-1/2}_{n}D^*_{n}(\beta)H^{-1/2}_{n}\lambda\ge \frac{1}{2}\Big)>1-\epsilon. \end{eqnarray} $

由向量值函数中值定理$ ^{[20]} $, 式(3.5), 引理3.7, 式(3.7)和(3.8) 可得, 以概率趋于1, 对一切$ \beta\in \partial \tilde B_n(r) $有

(3.9) $ \begin{eqnarray} &&(\beta-\beta_0)^\tau g_{n}(\beta) = (\beta-\beta_0)^\tau g_{n}-(\beta-\beta_0)^\tau[g_{n}-g_{n}(\beta)] \\ & = &(\beta-\beta_0)^\tau g_{n}-(\beta-\beta_0)^\tau D^*_{n}(\beta)(\beta-\beta_0)\\ & = &(\beta-\beta_0)^\tau\bar M_{n}\times\bar M^{-1}_{n} M^{1/2}_{n}\times M^{-1/2}_{n}g_{n}\\ &&-(\beta-\beta_0)^\tau H^{1/2}_{n}\left[ H^{-1/2}_{n}D^*_{n}(\beta)H^{-1/2}_{n}\right] H^{1/2}_{n}(\beta-\beta_0)\\ &\le&C r-\frac{C_5}{2}r^2. \end{eqnarray} $

因此, 取$ r $充分大使$ Cr-\frac{C_5}{2}r^2<0 $, 则有

(3.10) $ \begin{eqnarray} \mbox{P} \Big\{ \sup\limits_{\beta\in \partial \tilde B_n(r)}(\beta-\beta_0)^\tau g_{n}(\beta)<0\Big\}\to 1. \end{eqnarray} $

由式(3.10)和引理3.1知

(3.11) $ \mbox{P} \Big\{ \mbox{存在 }\ \hat\beta_{n}\in \tilde B_n(r)\ \mbox{ 使得 }\ g_{n}(\hat\beta_{n}) = 0\Big\}\to 1, $

所以, 由$ \hat\beta_{n}\in \tilde B_n(r) $, 式(3.5)和条件(A1)(ⅰ), 可得

(3.12) $ \begin{eqnarray} \|\hat\beta_{n}-\beta_0\| = \|F^{-1/2}_{n}.F^{1/2}_{n}(\bar M^\tau_{n})^{-1}.\bar M^\tau_{n}(\hat\beta_{n}-\beta_0)\| = o_p(1). \end{eqnarray} $

由式(3.11)和向量值函数中值定理^[20]知

(3.13) $ \begin{eqnarray} g_{n}(\beta_0) = D_{n}^*(\hat\beta_{n})(\hat\beta_{n}-\beta_0). \end{eqnarray} $

所以

(3.14) $ \begin{eqnarray} &&M_{n}^{-1/2}H_{n}(\hat\beta_{n}-\beta_0) \\& = & M_{n}^{-1/2} H_{n}^{1/2}\left\{[H_{n}^{-1/2}D^*_{n}(\hat\beta_{n})H_{n}^{-1/2}]^{-1}-I\right\}[M_{n}^{-1/2} H_{n}^{1/2}]^{-1}M_{n}^{-1/2}g_{n}(\beta_0) \\ &&+M_{n}^{-1/2}g_{n}(\beta_0). \end{eqnarray} $

由式(3.14), (3.3), (3.5)和引理3.7, 可得

(3.15) $ \begin{eqnarray} M_{n}^{-1/2}H_{n}(\hat\beta_{n}-\beta_0)\to_d N(0, I). \end{eqnarray} $

由式(3.11), (3.12) 和(3.15), 知定理2.1得证.

我们考虑相关的自适应设计logit模型:

$ \begin{eqnarray} E(Y_{ij}|{{\cal F}_{i-1}}) = \frac{\exp(\beta_{00}+\beta_{01}Y_{i-1, j})}{1+\exp(\beta_{00}+\beta_{01}Y_{i-1, j})}, \quad i = 1, \cdots , n, \quad j = 1, \cdots , m. \end{eqnarray} $

即$ \beta_0 = (\beta_{00}, \beta_{01})^\tau $, $ X_{ij} = (1, Y_{i-1, j})^\tau $, 其中$ Y_{0j} = 0 $. 首先我们取$ m = 3 $, $ \beta_0 = (-0.9, 0.4)^\tau $, $ Y_i = (Y_{i1}, Y_{i2}, Y_{i3})^\tau $有相关系数为0.3的交换相关结构(即$ \bar R_{i} = 0.3{\bf\bar 1_3}+0.7I_3 $, 其中$ \bf {\bar 1_3} $是元素都为1的$ 3\times 3 $矩阵, $ I_3 $是$ 3 $阶单位阵). 表 1给出了样本容量分别为$ n = 50, 100, 200, 400 $, 工作相关阵分别取真实相关阵($ R_i = \bar R_{i} $), 独立相关阵($ R_i = I $), 一阶自回归相关阵(即$ R_i $的第$ j $行, $ j' $列的元素为$ 0.3^{|j-j'|} $) 的模拟结果.

然后我们取$ m = 8 $, $ \beta_0 = (0.9, 0.4) $, $ Y_i = (Y_{i1}, Y_{i2}, \cdots, Y_{i8})^\tau $有相关系数为0.3的交换相关结构. 表 2给出了样本容量分别为$ n = 50, 100, 200, 400 $, 三种工作相关阵分别为真实相关阵, 独立(工作)相关阵, 一阶自回归(工作)相关阵的模拟结果.

从表 1和表 2可看出随着样本容量$ n $的增加, 回归参数估计的均值越来越接近真值, 偏差(标准差)越来越小. 而且当工作相关阵等于真实相关阵时, 估计的标准差最小.

引理3.4的证明  由假设(A1)(ⅰ), (3.5)式和$ \tilde B_n(r) $的定义知

(5.1) $ \begin{equation} \sup\limits_{i\le n, j, \beta\in \tilde B_n(r)}\| X_{ij}^\tau(\beta-\beta_0)\|\le \sup\limits_{i\le n, j, \beta\in \tilde B_n(r)} \| X_{ij}^\tau F_n^{-1/2}.(\bar M_n^{-1} F_n^{1/2})^\tau.\bar M_n^\tau(\beta-\beta_0)\| = o_p(1). \end{equation} $

由微分中值定理, 假设(A6)(ⅰ), 式(5.1)知

(5.2) $ \begin{eqnarray} \sup\limits_{i\le n, j, \beta\in \tilde B_n(r)}|A_{ij}(\beta)-A_{ij}| = |\mu^{(2)}(X_{ij}^\tau\beta^*)|\| X_{ij}^\tau(\beta-\beta_0)\| = o_p(1), \end{eqnarray} $

其中$ \beta^* $在$ \beta_0 $与$ \beta $的连线上. 由(A6)(ⅱ), 式(5.2)知

(5.3) $ \begin{eqnarray} \sup\limits_{i\le n, j, \beta\in \tilde B_n(r)}|A^{1/2}_{ij}(\beta)-A^{1/2}_{ij}| = \frac{|A_{ij}(\beta)-A_{ij}|}{A^{1/2}_{ij}(\beta)+A^{1/2}_{ij}} = o_p(1). \end{eqnarray} $

记$ \lambda_k $和$ \lambda_l $分别是第$ k $个和第$ l $个分量为1, 其余元素都为0的$ p $维单位向量, $ 1\le k, l\le p $. 由式(3.5)知

(5.4) $ \begin{eqnarray} \sum\limits_{i = 1}^n\lambda_{l}^\tau H^{-1/2}_nX_i X^\tau_i H^{-1/2}_n\lambda_{ l} = O_p(1). \end{eqnarray} $

由柯西-施瓦兹不等式, 式(5.3), (5.4), 条件(A4)和(A6)可得

(5.5) $ \begin{eqnarray} &&\sup\limits_{i\le n, \beta\in \tilde B_n(r)}\bigg(\sum\limits_{i = 1}^n\lambda_{k}^\tau H^{-1/2}_nX_i[ A^{ 1/2}_i(\beta)- A^{ 1/2}_i] R_i^{-1} A^{ 1/2}_i(\beta) X^\tau_i H^{-1/2}_n\lambda_{ l}\bigg)^2\\ &\le& \sup\limits_{i\le n, \beta\in \tilde B_n(r)}\sum\limits_{i = 1}^n\lambda_{ k}^\tau H^{-1/2}_nX_i[ A^{ 1/2}_i(\beta)- A^{ 1/2}_i]^2 X^\tau_i H^{-1/2}_n\lambda_{ k}\\ &&\times\sup\limits_{i\le n, \beta\in \tilde B_n(r)}\sum\limits_{i = 1}^n \lambda_{ l}^\tau H^{-1/2}_nX_i A^{ 1/2}_i(\beta)R_i^{-2}A^{ 1/2}_i(\beta) X^\tau_i H^{-1/2}_n\lambda_{ l}\\ &\le &\sup\limits_{i\le n, \beta\in \tilde B_n(r)} \lambda_{\max}\left\{[ A^{ 1/2}_i(\beta)- A^{ 1/2}_i ]^2 \right\}\sum\limits_{i = 1}^n\lambda_{k}^\tau H^{-1/2}_nX_i X^\tau_i H^{-1/2}_n\lambda_{ k}\\ &&\times \sup\limits_{i\ge 1, \beta\in \tilde B_n(r)}\lambda_{\max}(R_i^{-2} A_i (\beta))\sum\limits_{i = 1}^n\lambda_{l}^\tau H^{-1/2}_nX_i X^\tau_i H^{-1/2}_n\lambda_{ l}\\ & = &o_p(1). \end{eqnarray} $

同理

(5.6) $ \begin{eqnarray} \bigg( \sum\limits_{i = 1}^n\lambda_{ k}^\tau H^{-1/2}_nX_i A^{ 1/2}_iR_i^{-1}[A^{ 1/2}_i(\beta) -A^{ 1/2}_i ] X^\tau_i H^{-1/2}_n\lambda_{ l}\bigg)^2 = o_p(1). \end{eqnarray} $

由式(5.5), (5.6)和三角不等式知

(5.7) $ \begin{eqnarray} \sup\limits_{\beta\in \tilde B_n(r)}|\lambda_{ k}^\tau H^{-1/2}_n[H_n(\beta)-H_n] H^{-1/2}_n\lambda_{ l}| = o_p(1), \end{eqnarray} $

因而引理3.4成立.

引理3.5的证明  由微分中值定理, (A6)(ⅰ), 式(5.1)知

(5.8) $ \begin{eqnarray} \sup\limits_{i\le n, j, \beta\in \tilde B_n(r)}|\mu_{ij}(\beta)-\mu_{ij}| = |\mu^{(1)}(X_{ij}^\tau\beta^*)|| X_{ij}^\tau(\beta-\beta_0)| = o_p(1), \end{eqnarray} $

其中$ \beta^* $在$ \beta_0 $与$ \beta $的连线上. 由柯西-施瓦兹不等式, (A1)(ⅱ), (A4), (A6), 式(5.8)知

(5.9) $ \begin{eqnarray} &&\sup\limits_{i\le n, \beta\in \tilde B_n(r)}\lambda_{\max}\left\{\mbox{diag}[ R_i^{-1}A_i^{-1/2}(\beta)(\mu_i-\mu_i(\beta))] \right\}^2\\ &\le &\sup\limits_{i\ge 1, \beta\in \tilde B_n(r)}\lambda_{\max}\left\{ R_i^{-1}A_i^{-1}(\beta) R_i^{-1}]\right\}\times\sup\limits_{i\le n, \beta\in \tilde B_n(r)}\sum\limits_{j = 1}^{m_i}|\mu_{ij}-\mu_{ij}(\beta)|^2\\ & = & o_p(1), \end{eqnarray} $

(5.10) $ \begin{equation} \sup\limits_{i\ge 1, \beta\in \tilde B_n(r)}\lambda_{\max}\left\{A_i^{-1/2}(\beta)\ddot\mu_i(\beta)\right\}^2<\infty\quad a.s.. \end{equation} $

与证明(5.5)式同理, 由柯西-施瓦兹不等式, 式(5.10), (5.4), (5.4)可证

(5.11) $ \begin{eqnarray} \lambda_k^\tau B^{[1]}_n(\beta)\lambda_l = o_p(1). \end{eqnarray} $

由(A1)(ⅱ), (A4), (A6), 式(5.8)可得

(5.12) $ \begin{equation} \sup\limits_{i\le n, \beta\in \tilde B_n(r)}\lambda_{\max}\left\{A_i^{1/2}(\beta) R_i^{-1}[\mbox{diag}(\mu_i-\mu_i(\beta))]^2R_i^{-1}A_i^{1/2}(\beta) \right\} = o_p(1), \end{equation} $

(5.13) $ \begin{equation} \sup\limits_{i\ge 1\beta\in \tilde B_n(r)}\lambda_{\max}\left\{A_i^{-3/2}(\beta)\ddot\mu_i(\beta)\right\}^2<\infty \quad a.s. \end{equation} $

与证明(5.5)式同理, 由柯西-施瓦兹不等式, 式(5.13), (5.12), (5.4)可得

(5.14) $ \begin{eqnarray} \lambda_k^\tau B^{[2]}_n(\beta)\lambda_l = o_p(1) . \end{eqnarray} $

由式(5.11)和(5.14)知, 引理3.5成立.

引理3.6的证明  记

(5.15) $ \begin{equation} \bar{{\cal E}}^{[1]}_n(\beta) = \bar M_n^{-1}H_n^{1/2}{{\cal E}}^{[1]}_n(\beta)H_n^{1/2}(\bar M_n^{-1})^\tau, \end{equation} $

(5.16) $ \begin{equation} \bar{{\cal E}}^{[2]}_n(\beta) = \bar M_n^{-1}H_n^{1/2}{{\cal E}}^{[2]}_n(\beta)H_n^{1/2}(\bar M_n^{-1})^\tau, \end{equation} $

则可验证

(5.17) $ \begin{equation} \lambda_k^\tau\bar{{\cal E}}^{[1]}_n(\beta)\lambda_l = \sum\limits_{i = 1}^nS_{ni}^{[1]}+\sum\limits_{i = 1}^nS_{ni}^{[3]}+\sum\limits_{i = 1}^nS_{ni}^{[5]}, \end{equation} $

(5.18) $ \begin{equation} \lambda_k^\tau\bar{{\cal E}}^{[2]}_n(\beta)\lambda_l = \sum\limits_{i = 1}^nS_{ni}^{[2]}+\sum\limits_{i = 1}^nS_{ni}^{[4]}+\sum\limits_{i = 1}^nS_{ni}^{[6]}, \end{equation} $

其中

$ \begin{eqnarray*} && S_{ni}^{[1]} = \frac{-1}{2}\lambda_k^\tau M_n^{-1}X_iA_i^{-1/2} \ddot\mu_i \mbox{diag}(X_i^\tau(\bar M_n^{-1})^\tau\lambda_l) R_i^{-1}A_i^{-1/2}(Y_i-\mu_i), \\ &&S_{ni}^{[2]} = \frac{1}{2}\lambda_k^\tau\bar M_n^{-1}X_iA_i^{1/2} R_i^{-1}\mbox{diag}[ X_i^\tau (\bar M_n^{-1})^\tau\lambda_l]A_i^{-3/2}\ddot\mu_i(Y_i-\mu_i), \\ &&S_{ni}^{[3]} = \frac{-1}{2}\lambda_k^\tau\bar M_n^{-1}X_i[A_i^{-1/2}(\beta) \ddot\mu_i(\beta) -A_i^{-1/2} \ddot\mu_i ]\mbox{diag}(X_i^\tau (\bar M_n^{-1})^\tau\lambda_l) R_i^{-1}A_i^{-1/2}(\beta)(Y_i-\mu_i), \\ &&S_{ni}^{[4]} = \frac{1}{2}\lambda_k^\tau\bar M_n^{-1}X_i\{A_i^{1/2}(\beta)-A_i^{1/2} \}R_i^{-1}\mbox{diag}[ X_i^\tau (\bar M_n^{-1})^\tau\lambda_l]A_i^{-3/2}\ddot\mu_i(Y_i-\mu_i), \\ &&S_{ni}^{[5]} = \frac{-1}{2}\lambda_k^\tau\bar M_n^{-1}X_iA_i^{-1/2} \ddot\mu_i \mbox{diag}(X_i^\tau (\bar M_n^{-1})^\tau\lambda_l) R_i^{-1}[A_i^{-1/2}(\beta)-A_i^{-1/2}](Y_i-\mu_i), \\ && S_{ni}^{[6]} = \frac{1}{2} \lambda_k^\tau\bar M_n^{-1}X_iA_i^{1/2}(\beta) R_i^{-1}\mbox{diag}[ X_i^\tau (\bar M_n^{-1})^\tau\lambda_l]\{A_i^{-3/2}(\beta)\ddot\mu_i(\beta)-A_i^{-3/2}\ddot\mu_i\}(Y_i-\mu_i). \end{eqnarray*} $

由假设(A1)(ⅰ)和式(3.5), 可得

(5.19) $ \begin{equation} \max\limits_{i\le n}\|\lambda^\tau_l \bar M_n^{-1} X_i\|^2\le C \max\limits_{i\le n}\|F_n^{-1/2} X_i \|^2 = o_p(1), \end{equation} $

(5.20) $ \begin{equation} \sum\limits_{i = 1}^n\|\lambda^\tau_k \bar M_n^{-1} X_i\|^2\le\sum\limits_{i = 1}^n\| \bar M_n^{-1} X_i\|^2 = O_p(1). \end{equation} $

由假设(A4)和(A6), 知

(5.21) $ \begin{eqnarray} \sup\limits_{i\ge 1} \lambda_{\max}(R_i^{-1}\bar R_i R_i^{-1})<\infty, a.s., \quad\sup\limits_{i\ge 1}\lambda_{\max}(A_i^{-1}\ddot\mu^2_i))<\infty, a.s. \end{eqnarray} $

由式(5.19), (5.20)和(5.21), 可得

(5.22) $ \begin{eqnarray} \sum\limits_{i = 1}^nE[(S_{ni}^{[1]})^2|{\cal F}_{i-1}] & = &\frac{1}{4} \sum\limits_{i = 1}^n\lambda_k^\tau\bar M_n^{-1}X_iA_i^{-1/2} \ddot\mu_i\mbox{diag}(\lambda_l^\tau \bar M_n^{-1}X_i) R_i^{-1}\bar R_i R_i^{-1}{}\\ & &{\qquad}\ \times\mbox{diag}(\lambda_l^\tau\bar M_n^{-1}X_i)\ddot\mu_i A_i^{-1/2}X_i^\tau(\bar M_n^{-1})^\tau\lambda_k{}\\ &\le &\frac{1}{4}\max\limits_{1\le i\le n} \lambda_{\max}(R_i^{-1}\bar R_i R_i^{-1})\times o_p(1) \times\max\limits_{1\le i\le n}\lambda_{\max}(A_i^{-1}\ddot\mu^2_i) \times O_p(1){}\\ & = &o_p(1). \end{eqnarray} $

所以, 对任意$ \epsilon>0 $, 也有

(5.23) $ \begin{eqnarray} \sum\limits_{i = 1}^nE[(S_{ni}^{[1]})^2(|S_{ni}^{[1]}|>\epsilon)|{\cal F}_{i-1}] = o_p(1). \end{eqnarray} $

由式(5.22), (5.23)和引理3.2, 取$ \eta = 0 $, 知

(5.24) $ \begin{equation} \sum\limits_{i = 1}^nS_{ni}^{[1]} = o_p(1). \end{equation} $

同理

(5.25) $ \begin{equation} \sum\limits_{i = 1}^nS_{ni}^{[2]} = o_p(1). \end{equation} $

(5.26) $ \begin{eqnarray} \sup\limits_{\beta\in \tilde B_n(r)}|\sum\limits_{i = 1}^nS_{ni}^{[3]} | &\le& \sum\limits_{i = 1}^n\sup\limits_{\beta\in \tilde B_n(r)}\|\frac{-1}{2}\lambda_k^\tau\bar M_n^{-1}X_i[A_i^{-1/2}(\beta) \ddot\mu_i(\beta) -A_i^{-1/2} \ddot\mu_i ]\\ &&{\qquad}{\qquad}{\quad}\times\mbox{diag}(X_i^\tau (\bar M_n^{-1})^\tau\lambda_l) R_i^{-1}A_i^{-1/2}(\beta)\|\| Y_i-\mu_i \|\\ &\le&\delta_n\times \sum\limits_{i = 1}^n\|\frac{-1}{2}\lambda_k^\tau\bar M_n^{-1}X_i\|\times\|X_i^\tau (\bar M_n^{-1})^\tau\lambda_l\| \times\| Y_i-\mu_i \|\\ &\le&\delta_n\times\sum\limits_{i = 1}^n\|\bar M_n^{-1}X_i\|^2 \times\| Y_i-\mu_i \|, \end{eqnarray} $

其中$ \delta_n = \sup\limits_{i\le n, \beta\in \tilde B_n(r)} \|A_i^{-1/2}(\beta) \ddot\mu_i(\beta) -A_i^{-1/2} \ddot\mu_i\|\times\sup\limits_{\beta\in \tilde B_n(r)} \|R_i^{-1}A_i^{-1}(\beta)\| $.

由微分中值定理, 假设(A4), (A6)和式(5.1)可得

(5.27) $ \begin{eqnarray} \delta_n = o_p(1). \end{eqnarray} $

由假设(A3), (A6)(ⅰ), 式(5.20)得

(5.28) $ \begin{equation} \sum\limits_{i = 1}^n\|\bar M_n^{-1}X_i\|^2E[\|Y_i-\mu_i\||{\cal F}_{i-1}]\le \sup\limits_{i\ge 1}E[\|Y_i-\mu_i\||{\cal F}_{i-1}]\times\sum\limits_{i = 1}^n\|\bar M_n^{-1}X_i\|^2 = O_p(1). \end{equation} $

记$ e_i = \|Y_i-\mu_i\|-E[\|Y_i-\mu_i\||{\cal F}_{i-1}] $. 由假设(A3), (A6)(ⅰ), Jesen不等式, 式(5.19), (5.20)可得

(5.29) $ \begin{equation} \sum\limits_{i = 1}^nE[\|\bar M_n^{-1}X_i\|^4e_i^2|{\cal F}_{i-1}] \le E[e_i^2|{\cal F}_{i-1}]\times\max\limits_{i\le n} \|\bar M_n^{-1}X_i\|^{2}\times\sum\limits_{i = 1}^n\|\bar M_n^{-1}X_i\|^2 = o_p(1), \end{equation} $

所以也有

(5.30) $ \begin{eqnarray} && \sum\limits_{i = 1}^nE[\|\bar M_n^{-1}X_i\|^4e_i^2I(\|\bar M_n^{-1}X_i\|^2|e_i|>\epsilon)|{\cal F}_{i-1}] = o_p(1). \end{eqnarray} $

由引理3.2, 式(5.29), (5.30), 取$ \eta = 0 $, 则有

(5.31) $ \begin{eqnarray} \sum\limits_{i = 1}^n\|\bar M_n^{-1}X_i\|^2e_i = o_p(1). \end{eqnarray} $

由式(5.26), (5.27), (5.28)和(5.31)知

(5.32) $ \begin{eqnarray} \sup\limits_{\beta\in \tilde B_n(r)}|\sum\limits_{i = 1}^nS_{ni}^{[3]} | = o_p(1). \end{eqnarray} $

同理可证

(5.33) $ \begin{eqnarray} \sup\limits_{\beta\in \tilde B_n(r)}|\sum\limits_{i = 1}^nS_{ni}^{[4]} | = o_p(1), \quad \sup\limits_{\beta\in \tilde B_n(r)}|\sum\limits_{i = 1}^nS_{ni}^{[5]} | = o_p(1) , \quad \sup\limits_{\beta\in \tilde B_n(r)}|\sum\limits_{i = 1}^nS_{ni}^{[6]} | = o_p(1). \end{eqnarray} $

由式(5.33), (5.32), (5.25), (5.24), (5.18), (5.17), (5.16), (5.15)和(3.5), 知引理3.6成立.

引理3.7的证明  对任意单位向量$ \lambda $, 记$ a_{ni} = \lambda^\tau\bar M_n^{-1}X_iA_i^{1/2}R_i^{-1}, Z_{ni} = a_{ni}Y^*_i $,

由(A1), (A4), (A6)和(3.5)式, 知

(5.34) $ \begin{equation} \max\limits_{i\le n}\|a_{ni}\|\le \|\lambda^\tau\bar M_n^{-1} F_n^{1/2}\| \|\max\limits_{i\le n}\|F_n^{-1/2} X_i \|\|A_i^{1/2}R_i^{-1}\| = o_p(1), \end{equation} $

(5.35) $ \begin{equation} \sum\limits_{i = 1}^n\|a_{ni}\|^2\le [\inf\limits_i\lambda_{\min}(\bar R_i)]^{-1}\sum\limits_{i = 1}^n\lambda^\tau\bar M_n^{-1}M_n(\bar M_n^{-1})^\tau\lambda^\tau = O_p(1). \end{equation} $

对任意$ \epsilon>0 $, 由式(5.34), (5.35)和假设(A3), 知

(5.36) $ \begin{equation} \sum\limits_{i = 1}^nE[Z_{ni}^2(I(|Z_{ni}|>\epsilon))|{{\cal F}}_{i-1}]\le\sup\limits_{i\ge 1} E(\|Y^*_i\|^{\alpha_0}|{{\cal F}}_{i-1}) \max\limits_{i\le n}\|a_{ni}\|^{\alpha_0-2}\sum\limits_{i = 1}^n\|a_{ni}\|^2\to_p0. \end{equation} $

由假设(A5), 知

(5.37) $ \begin{equation} \sum\limits_{i = 1}^nE[Z_{ni}^2|{{\cal F}}_{i-1}] = \lambda^\tau\bar M_n^{-1}M_n(\bar M_n^{-1})^\tau\lambda\to_p 1. \end{equation} $

由假设(A5), 式(5.36), (5.37)和引理3.2, 知

(5.38) $ \begin{eqnarray} M_n^{-1/2}g_n(\beta_0) = [\bar M^{-1}_n M_n^{1/2}]^{-1} \times\bar M_n^{-1}g_n(\beta_0) \to_d N(0, I), \end{eqnarray} $

引理3.7得证.